Hachiman Wake Shrine (1890), Sako, Akaiwa City, Okayama Prefecture (17)


Problem

When six spheres with a diameter of $8 \ cm$ are arranged on a flat surface as shown in the figure, and a sphere with a diameter of $10 \ cm$ is placed on top of them, find the height of this object.


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Solution

If you draw a three-dimensional figure faithfully to the problem statement, it will look like the following:

 

Let the diameters of the six spheres in the bottom row be $x$, and the diameter of the sphere in the top row be $y$.
Here, $A$ is the vertex of this object, $B$ is the center of the regular hexagon formed by connecting the centers of the six spheres in the lower row, $C$ is the foot of the perpendicular line drawn from $A$ to the plane, and $D$ is the tangent point of the spheres centered on $O_1$ and $O_2$, respectively.
If you look at the bottom six spheres from above, it will look like the figure below:

Therefore, the following can be said:

$$BD=\frac{\sqrt{3}}{2} x.$$

Regarding the right triangle $O_7 O_1D$ in the solid figure above, we know from the Pythagorean theorem that

$$DO_7^2+O_1 D^2=O_7 O_1^2,$$

$$DO_7^2+(\frac{x}{2})^2=(\frac{x+y}{2})^2,$$

$$DO_7^2=\frac{xy}{2}+\frac{y^2}{4},$$

$$∴ \ DO_7=\sqrt{\frac{xy}{2}+\frac{y^2}{4}}.$$

Therefore, for the right triangle $O_7 BD$, the following holds:

$$O_7B^2+BD^2=DO_7^2,$$

$$O_7B^2+(\frac{\sqrt{3}}{2} x)^2=(\sqrt{\frac{xy}{2}+\frac{y^2}{4}})^2,$$

$$O_7B^2=\frac{y^2+2xy-3x^2}{4},$$

$$∴ \ O_7B=\frac{\sqrt{y^2+2xy-3x^2}}{2}.$$

Because the height of this object is the length of $AC=AO_7+O_7B+BC$, we know that

$$AC=\frac{x+y+\sqrt{y^2+2xy-3x^2}}{2}. \qquad [*]$$

Since $x=8$ and $y=10$ from the problem statement, by substituting these into $[*]$, we have the following:

$$AC=\frac{8+10+\sqrt{10^2+2×8×10-3×8^2}}{2}=\frac{18+\sqrt{68}}{2}=9+\sqrt{17}≒13.123.$$

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$$Ans.\ 13.123 \ cm.\quad $$


Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.52; pp.302-303.